The sum of the series frac{3}{{1! + 2! + 3!}} | vedclass

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(C) The general term of the series is $T_n = \frac{n+2}{n! + (n+1)! + (n+2)!}$ for $n = 1, 2, \dots, 2006$.$T_n = \frac{n+2}{n! [1 + (n+1) + (n+2)(n+1

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$\frac{{\left( {2008} \right)! - 2}}{{2.\left( {2008} \right)!}}$

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(C) The general term of the series is $T_n = \frac{n+2}{n! + (n+1)! + (n+2)!}$ for $n = 1, 2, \dots, 2006$.$T_n = \frac{n+2}{n! [1 + (n+1) + (n+2)(n+1)]}$$T_n = \frac{n+2}{n! [n+2 + (n+2)(n+1)]} = \frac{n+2}{n! (n+2) [1 + n + 1]} = \frac{n+2}{n! (n+2) (n+2)} = \frac{1}{n! (n+2)}$This does not simplify to a simple telescoping series directly. Let us re-evaluate the original series term: $T_n = \frac{n+2}{n! + (n+1)! + (n+2)!}$.For $n=1$,$T_1 = \frac{3}{1!+2!+3!} = \frac{3}{1+2+6} = \frac{3}{9} = \frac{1}{3}$.Using the telescoping form $T_n = \frac{1}{n!} - \frac{1}{(n+2)!}$ is incorrect. Let us use $T_n = \frac{n+2}{n! (1 + n+1 + (n+2)(n+1))} = \frac{n+2}{n! (n+2)^2} = \frac{1}{n!(n+2)}$.Actually,the standard approach for this series is $T_n = \frac{n+2}{n!(1 + (n+1) + (n+1)(n+2))} = \frac{n+2}{n!(n+2)^2} = \frac{1}{n!(n+2)}$.Summing this from $n=1$ to $2006$ gives $\sum_{n=1}^{2006} \frac{1}{n!(n+2)} = \sum_{n=1}^{2006} \frac{n+1}{(n+2)!} = \sum_{n=1}^{2006} \left( \frac{n+2-1}{(n+2)!} \right) = \sum_{n=1}^{2006} \left( \frac{1}{(n+1)!} - \frac{1}{(n+2)!} \right)$.This is a telescoping sum: $(\frac{1}{2!} - \frac{1}{3!}) + (\frac{1}{3!} - \frac{1}{4!}) + \dots + (\frac{1}{2007!} - \frac{1}{2008!}) = \frac{1}{2} - \frac{1}{2008!} = \frac{2008! - 2}{2 \cdot 2008!}$.

The sum of the series $\frac{3}{{1! + 2! + 3!}} + \frac{4}{{2! + 3! + 4!}} + \frac{5}{{3! + 4! + 5!}} + ...... + \frac{{2008}}{{\left( {2006} \right)! + \left( {2007} \right)! + \left( {2008} \right)!}}$ is equal to

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